What is the integral of (2x+1)/((3x-1)(x^2+2x+2)) ?

The integral of (2x+1)/((3x-1)(x^2+2x+2)) is 1/5 ln|3x-1|+1/10 (-ln|x^2+2x+2|+4arctan(x+1))+C

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integral of (2x+1)/((3x-1)(x^2+2x+2))

\int \frac{2 x + 1}{( 3 x - 1 ) ( x ^{2} + 2 x + 2 )} d x

\frac{1}{5} ln ∣ 3 x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 4 arctan (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{( 3 x - 1 ) ( x ^{2} + 2 x + 2 )} d x

Take the partial fraction of

\frac{2 x + 1}{( 3 x - 1 ) ( x ^{2} + 2 x + 2 )} : \frac{3}{5 ( 3 x - 1 )} + \frac{- x + 1}{5 ( x ^{2} + 2 x + 2 )}

= \int \frac{3}{5 ( 3 x - 1 )} + \frac{- x + 1}{5 ( x ^{2} + 2 x + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{5 ( 3 x - 1 )} d x + \int \frac{- x + 1}{5 ( x ^{2} + 2 x + 2 )} d x

\int \frac{3}{5 ( 3 x - 1 )} d x = \frac{1}{5} ln ∣ 3 x - 1 ∣

\int \frac{- x + 1}{5 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{10} (- ln x^{2} + 2 x + 2 + 4 arctan (x + 1))

= \frac{1}{5} ln ∣ 3 x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 4 arctan (x + 1))

Add a constant to the solution

= \frac{1}{5} ln ∣ 3 x - 1 ∣ + \frac{1}{10} (- ln x^{2} + 2 x + 2 + 4 arctan (x + 1)) + C

\frac{\partial}{\partial x} (4 - 2 \cdot (3 x - y)^{2})

y = - (- 6 x + 6)^{\frac{1}{2}}, x = - 5

\frac{d}{d v} (a e^{v} + \frac{b}{v} + \frac{c}{v ^{2}})

x y^{^{'}} - 2 y + y^{2} = 0

y^{^{''}} + 2 k y^{^{'}} + k^{4} y = 0

What is the integral of (2x+1)/((3x-1)(x^2+2x+2)) ?
The integral of (2x+1)/((3x-1)(x^2+2x+2)) is 1/5 ln|3x-1|+1/10 (-ln|x^2+2x+2|+4arctan(x+1))+C